Chapter 4.5, Problem 48PS

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

Chapter
Section

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

# For Problems 11-52, perform the indicated divisions. (Objective 1) ( x 5 + 1 ) ÷ ( x + 1 )

To determine

To Find:

The value of the polynomials by performing the indicated division by binomials.

Explanation

Division of a polynomial by binomial is done by using the long-division method.

In this method to find the first term of the quotient, divide the first term of the dividend by the first term of the divisor.

Multiply the divisor with the term obtained by dividing the first term of the dividend with the first term of the quotient.

The product obtained is subtracted from the dividend.

Repeat the same process to divide the other terms of the dividend.

Calculation:

Consider the expression (x5+1)Ã·(x+1).

The following steps are used to solve the division problem.

First multiply the divisor by x4 and write the product x5+x4 under the dividend and subtract.

The value obtained is equals to âˆ’x4Â .

Now multiply the divisor by âˆ’x3 and write the product âˆ’x4âˆ’x3 under the dividend and

Subtract.

The value obtained is equal to Â x3.

Now multiply the divisor by x2 and write the product x3+x2 under the dividend and

Subtract.

The value obtained is equal to âˆ’x2.

Now multiply the divisor by âˆ’x and write the product âˆ’x2âˆ’xÂ  under the dividend and

Subtract.

The value obtained is equal to x.

Now multiply the divisor by 1 and write the product x+1 under the dividend and simplify.

Here write the missing exponents as 0x4,0x3,0x2,0x and solve the division.

By using the long division it is written as follows,

x+1x4âˆ’x3+x2âˆ’x+1x5+0x4+0x3+0x2+0x+1Â Â Â Â Â Â Â Â x5+x4Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â _Â Â Â Â Â Â Â Â Â Â Â Â Â Â âˆ’x4Â +0x3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â âˆ’x4âˆ’x3Â Â Â Â Â Â Â Â Â Â Â Â Â _Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x3+0x2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x3+x2Â Â Â Â Â Â Â Â Â Â Â Â Â _Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â âˆ’x2+0xÂ Â

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